Tính:
a)\(\frac{{ - 1}}{6} + 0,75\);
b)\(3\frac{1}{{10}} - \frac{3}{8}\);
c)\(0,1 + \frac{{ - 9}}{{17}} - \left( { - 0,9} \right)\).
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HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)
c)
\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)
d)
\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)
e)
\(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)
g)
\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)
15 tháng 8 2023
a: \(\left(\dfrac{1}{5}\right)^{-2}=25\)
b: \(4^{\dfrac{3}{2}}=8\)
c: \(\left(\dfrac{1}{8}\right)^{-\dfrac{2}{3}}=\left(\dfrac{1}{2}\right)^{3\cdot\dfrac{-2}{3}}=\left(\dfrac{1}{2}\right)^{-2}=4\)
d: \(\left(\dfrac{1}{16}\right)^{-0.75}=\left(\dfrac{1}{2}\right)^{4\cdot\left(-0.75\right)}=\left(\dfrac{1}{2}\right)^{-3}=8\)
NH
4
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);
b)\({\left( {0,75 - 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{6}{4}} \right)^3} = {\left( { - \frac{3}{4}} \right)^3} = \frac{{ - 27}}{{64}};\)
c)
\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)
d) \({\left( {1 - \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} - \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)
b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)
c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)
d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)
e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)
g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)
i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)
k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)
15 tháng 8 2023
a: \(=3^2+\left(3^4\right)^{-0.75}-5^{2\cdot0.5}\)
\(=9-5+3^{-3}=4+\dfrac{1}{27}=\dfrac{109}{27}\)
b: \(=2^{4-6\sqrt{7}}\cdot2^{6\sqrt{7}}=2^{4-6\sqrt{7}+6\sqrt{7}}=2^4=16\)
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
\(\begin{array}{l}a)\frac{{ - 6}}{{18}} + \frac{{18}}{{27}}\\ = \frac{{ - 1}}{3} + \frac{2}{3}\\ = \frac{1}{3}\\b)2,5 - ( - \frac{6}{9})\\ = \frac{{25}}{{10}} + \frac{6}{9}\\ = \frac{5}{2} + \frac{2}{3}\\ = \frac{{15}}{6} + \frac{4}{6}\\ = \frac{{19}}{6}\\c) - 0,32.( - 0,875)\\ = \frac{{ - 32}}{{100}}.( - \frac{{875}}{{1000}})\\ = \frac{{ - 8}}{{25}}.(\frac{{ - 7}}{8})\\ = \frac{8}{{25}}.\frac{7}{8}\\ = \frac{7}{{25}}\\d)( - 5):2\frac{1}{5}\\ = ( - 5):\frac{{11}}{5}\\ = ( - 5).\frac{5}{{11}}\\ = \frac{{ - 25}}{{11}}\end{array}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
a)\(\frac{{ - 1}}{6} + 0,75 = \frac{{ - 1}}{6} + \frac{3}{4} = \frac{{ - 2}}{{12}} + \frac{9}{{12}} = \frac{7}{{12}}\);
b)\(3\frac{1}{{10}} - \frac{3}{8} = \frac{{31}}{{10}} - \frac{3}{8} = \frac{{124}}{{40}} - \frac{{15}}{{40}} = \frac{{109}}{{40}}\);
c)
\(\begin{array}{l}0,1 + \frac{{ - 9}}{{17}} - \left( { - 0,9} \right) = \frac{1}{{10}} + \frac{{ - 9}}{{17}} + \frac{9}{{10}}\\ = (\frac{1}{{10}} + \frac{9}{{10}}) + \frac{{ - 9}}{{17}} = 1 + \frac{{ - 9}}{{17}} =\frac{{ 17}}{{17}}+\frac{{ - 9}}{{17}}= \frac{8}{{17}}\end{array}\)